Question

An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fluid flow through the 3.0-inch section will be what factor times that through the 1.0-inch section

Answer 1

Answer:

$(\frac{r_1}{r_2})^2=\frac{1}{9}$

Explanation:

From the question we are told that:

Diameter 1 $d_1=1.0$

Diameter 2 $d_2=3.0$

Generally the equation for Radius is mathematically given by

At Diameter 1

$r_{1}=\frac{1}{2} inch$

At Diameter 2

$r_{2}=\frac{3}{2} inch$

Generally the equation for continuity is mathematically given by

 $A_1V_1=A_2V_2$

Therefore

$(\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2$

$(\frac{r_1}{r_2})^2=\frac{1}{9}$